A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
The x coordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?
anonymous
 3 years ago
The x coordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.09y^2=x^3 d(9y^2)/dx =d(x^3)/dx 18y dy/dx =3x^2 dy/dx =x^2/(6y)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The gradient of tangent is x^2/(6y) So, gradient of normal is 6y/(x^2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0NOW, let xintercept of the normal be a then yintercept is also a.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Gradient of normal = (a0)/(0a) 6y/(x^2) =1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now,dw:1359201825424:dw and solve for x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359201971611:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359202095063:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.06y/(x^2) =1......then wat did u do ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Substitute y interms of x
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.